Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m, q, fa[100005][18], dep[100005];
vector<int> E[100005];
struct hzl {
  int id[15], n;
  hzl() { n = 0; }
  hzl operator+(const hzl &x) {
    hzl ans;
    int l1 = 0, l2 = 0, s = 0;
    while (l1 < n && l2 < x.n && s < 10) {
      if (id[l1] < x.id[l2])
        ans.id[s++] = id[l1++];
      else
        ans.id[s++] = x.id[l2++];
    }
    while (l1 < n && s < 10) ans.id[s++] = id[l1++];
    while (l2 < x.n && s < 10) ans.id[s++] = x.id[l2++];
    ans.n = s;
    return ans;
  }
  void Print(int v) {
    int mi = min(n, v);
    printf("%d", mi);
    for (int i = 0; i < mi; i++) printf(" %d", id[i]);
    puts("");
  }
} A[100005][18];
void dfs(int x, int f) {
  fa[x][0] = f;
  for (int i = 0; i < E[x].size(); i++) {
    int y = E[x][i];
    if (y == f) continue;
    dep[y] = dep[x] + 1;
    dfs(y, x);
  }
}
void UP(int &x, int y, hzl &ans) {
  for (int i = 0; i <= 17; i++)
    if (y & (1 << i)) {
      ans = ans + A[x][i];
      x = fa[x][i];
    }
}
void solve(int a, int b, int v) {
  if (a == b) {
    A[a][0].Print(v);
    return;
  }
  hzl ans;
  if (dep[a] > dep[b]) swap(a, b);
  UP(b, dep[b] - dep[a], ans);
  if (a == b) {
    ans = ans + A[a][0];
    ans.Print(v);
    return;
  }
  for (int i = 17; i >= 0; i--)
    if (fa[a][i] != fa[b][i]) {
      ans = ans + A[a][i] + A[b][i];
      a = fa[a][i];
      b = fa[b][i];
    }
  ans = ans + A[a][1] + A[b][0];
  ans.Print(v);
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    E[a].push_back(b);
    E[b].push_back(a);
  }
  for (int i = 1; i <= m; i++) {
    int a;
    scanf("%d", &a);
    if (A[a][0].n < 10) A[a][0].id[A[a][0].n++] = i;
  }
  dfs(1, 0);
  for (int i = 1; i <= 17; i++) {
    for (int j = 1; j <= n; j++) {
      fa[j][i] = fa[fa[j][i - 1]][i - 1];
      A[j][i] = A[fa[j][i - 1]][i - 1] + A[j][i - 1];
    }
  }
  for (int i = 1; i <= q; i++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    solve(a, b, c);
  }
  return 0;
}